It is quite easy to find palindrome replacing letters on both ends with (Φ).
ΦabaΦ
ΦΦbaΦ
ΦΦbaΦ
ΦΦbaΦ
ΦΦbaΦ
ΦΦbaΦ
ΦΦbΦΦ
ΦΦbΦΦ
ΦΦbΦΦ
ΦΦΦΦΦ
ΦΦΦΦΦ
ΦΦΦΦΦ YES
It is possible with changing a to A , and changing A to a at the end of task. But does anybody have an idea ,how to achieve this without using additional signs?
Moving of end characters can help:
ΦΦabaΦΦ
ΦaΦbaΦΦ
ΦaΦbΦaΦ YES
or
ΦΦabbaΦΦ
ΦaΦbbaΦΦ
ΦaΦbbΦaΦ
ΦabΦbΦaΦ
ΦabΦΦbaΦ YES
You can change a to A, and b to B, and ur end character would be a capital letter. (take the ascia value, and if it falls in the uppercase range, you know the end)
This assume your input is entirely lower case.
For each possible symbol on the tape (which I assume are drawn from a finite set), you need a state, which I'll call "$X_LOOKING". Start at the left end and set the state to "$X_LOOKING" for the symbol $X that you find there. Move right until you hit the end and see if it matches $X.
When you move back left, you'll have to stop at the 2nd letter instead of the first. For this maybe you can keep track of how many letters you've looked at on another area of the tape.
Related
Im new so if this question was already Asked (i didnt find it scrolling through the list of results though) please send me the link.
I got a math quiz and im to lazy to go through all the possibilities so i thought i can find a program instead. I know a bit about programming but not much.
Is it possible (and in what programming language, and how) to read only one digit, e.g at the 3rd Position, in a integer?
And how is an integer actually saved, in a kind of array?
Thanks!
You can get rid of any lower valued digit (the ones and tens if you only want the hundreds) by dividing with rounding/truncation. 1234/100 is 12 in most languages if you are doing integer division.
You can get rid of any higher valued digits by using taking the modulus. 12 % 10 is 2 in many languages; just find out how the modulus is done in yours. I use "modulus" meaning "divide and keep the rest only", i.e. it is the opposite of "divide with rounding"; that which is lost by rounding is the final result of the modulus.
The alternative is however to actually NOT see the input as a number and treat it as text. That way it is often easier to ignore the last 2 characters and all leading characters.
could anyone tell me that
is it right way if i calculate the crc by the method given below:
i did this because dataword length is large.
is it right?
The bottom set of divisions are what you want for a CRC, and arrive at the correct remainders. Your quotients each need one more bit, but they are not used anyway.
The top division was not completed, but it is not relevant, since you need to append the four zeros first as you did in the bottom left, or the four-bit CRC as you did in the bottom right.
Ultimately, You are doing the same thing what a division does. Refer https://www.wikihow.com/Divide-Binary-Numbers binary division for more. However, the data word to be sent to the receiver should not be altered.
What's the purpose of array indices starting at 0 in most programming languages, in contrast to the ordinal way in which we refer to most things IRL (first, second, third, etc.)? What's the logic or utility behind that?
I'm completely used to it by now, but never stopped to think about the reason behind it.
Update: One benefit I read about from Googling is that for loops can have i < n if you want to go up to n.
Dijkstra lays out the reasoning in Why numbering should start at zero.
When dealing with a sequence of length N, the elements of which we wish to distinguish by subscript, the next vexing question is what subscript value to assign to its starting element...
when starting with subscript 1, the subscript range 1 ≤ i < N+1; starting with 0, however, gives the nicer range 0 ≤ i < N. So let us let our ordinals start at zero: an element's ordinal (subscript) equals the number of elements preceding it in the sequence. And the moral of the story is that we had better regard —after all those centuries!— zero as a most natural number.
When we're accessing item by index like a[i] the compiler converts it to [a+i]. So the index of first element is zero because [a+0] will give us 'a' that points to the first item in array. This is quite obviously for, say, C++ but not for more recent languages such as C#.
Dijkstra wrote a really interesting paper about this, in 1982: Why numbering should start at zero.
You may google for it, there's been a lot of discussions about it. I'd say that the fact that the offset of the first element from the beginning (which it is) is zero certainly makes sense.
Because Dijkstra said so.
In my old assembler days it was natural for the offset to start at zero.
dcl foo(9)
ldx0 0 'offset to index register 0
lda foo,x0 'get first element
adx0 1,du 'get 2nd
ldq foo,x0
When looking at it from the perspective of the hardware it makes more sense.
can any one please explain why this gives different outputs?
round(1.49999999999999)
1
round(1.4999999999999999)
2
I have read the round documentation but it does not mention anything about it there.
I know that R represents numbers in binary form, but why does adding two extra 9's changes the result?
Thanks.
1.4999999999999999 can't be represented internally, so it gets rounded to 1.5.
Now, when you apply round(), the result is 2.
Put those two numbers into variable and then print it - you'll see they are different.
Computers doesn't store this kind of numbers with this exact value, (They don't use decadic numbers internaly)
I have never used R, so I don't know is this is the issue, but in other languages such as C/C++ a number like 1.4999999999999999 is represented by a float or a double.
Since these have finite precision, you cannot represent something like 1.4999999999999999 exactly. It might be the case that 1.4999999999999999 actually gets stored as 1.50000000000000 instead due to limitations on floating point precision.
I have been playing with an implementation of lookandsay (OEIS A005150) in J. I have made two versions, both very simple, using while. type control structures. One recurs, the other loops. Because I am compulsive, I started running comparative timing on the versions.
look and say is the sequence 1 11 21 1211 111221 that s, one one, two ones, etc.
For early elements of the list (up to around 20) the looping version wins, but only by a tiny amount. Timings around 30 cause the recursive version to win, by a large enough amount that the recursive version might be preferred if the stack space were adequate to support it. I looked at why, and I believe that it has to do with handling intermediate results. The 30th number in the sequence has 5808 digits. (32nd number, 9898 digits, 34th, 16774.)
When you are doing the problem with recursion, you can hold the intermediate results in the recursive call, and the unstacking at the end builds the results so that there is minimal handling of the results.
In the list version, you need a variable to hold the result. Every loop iteration causes you to need to add two elements to the result.
The problem, as I see it, is that I can't find any way in J to modify an extant array without completely reassigning it. So I am saying
try. o =. o,e,(0&{y) catch. o =. e,(0&{y) end.
to put an element into o where o might not have a value when we start. That may be notably slower than
o =. i.0
.
.
.
o =. (,o),e,(0&{y)
The point is that the result gets the wrong shape without the ravels, or so it seems. It is inheriting a shape from i.0 somehow.
But even functions like } amend don't modify a list, they return a list that has a modification made to it, and if you want to save the list you need to assign it. As the size of the assigned list increases (as you walk the the number from the beginning to the end making the next number) the assignment seems to take more time and more time. This assignment is really the only thing I can see that would make element 32, 9898 digits, take less time in the recursive version while element 20 (408 digits) takes less time in the loopy version.
The recursive version builds the return with:
e,(0&{y),(,lookandsay e }. y)
The above line is both the return line from the function and the recursion, so the whole return vector gets built at once as the call gets to the end of the string and everything unstacks.
In APL I thought that one could say something on the order of:
a[1+rho a] <- new element
But when I try this in NARS2000 I find that it causes an index error. I don't have access to any other APL, I might be remembering this idiom from APL Plus, I doubt it worked this way in APL\360 or APL\1130. I might be misremembering it completely.
I can find no way to do that in J. It might be that there is no way to do that, but the next thought is to pre-allocate an array that could hold results, and to change individual entries. I see no way to do that either - that is, J does not seem to support the APL idiom:
a<- iota 5
a[3] <- -1
Is this one of those side effect things that is disallowed because of language purity?
Does the interpreter recognize a=. a,foo or some of its variants as a thing that it should fastpath to a[>:#a]=.foo internally?
This is the recursive version, just for the heck of it. I have tried a bunch of different versions and I believe that the longer the program, the slower, and generally, the more complex, the slower. Generally, the program can be chained so that if you want the nth number you can do lookandsay^: n ] y. I have tried a number of optimizations, but the problem I have is that I can't tell what environment I am sending my output into. If I could tell that I was sending it to the next iteration of the program I would send it as an array of digits rather than as a big number.
I also suspect that if I could figure out how to make a tacit version of the code, it would run faster, based on my finding that when I add something to the code that should make it shorter, it runs longer.
lookandsay=: 3 : 0
if. 0 = # ,y do. return. end. NB. return on empty argument
if. 1 ~: ##$ y do. NB. convert rank 0 argument to list of digits
y =. (10&#.^:_1) x: y
f =. 1
assert. 1 = ##$ y NB. the converted argument must be rank 1
else.
NB. yw =. y
f =. 0
end.
NB. e should be a count of the digits that match the leading digit.
e=.+/*./\y=0&{y
if. f do.
o=. e,(0&{y),(,lookandsay e }. y)
assert. e = 0&{ o
10&#. x: o
return.
else.
e,(0&{y),(,lookandsay e }. y)
return.
end.
)
I was interested in the characteristics of the numbers produced. I found that if you start with a 1, the numerals never get higher than 3. If you start with a numeral higher than 3, it will survive as a singleton, and you can also get a number into the generated numbers by starting with something like 888888888 which will generate a number with one 9 in it and a single 8 at the end of the number. But other than the singletons, no digit gets higher than 3.
Edit:
I did some more measuring. I had originally written the program to accept either a vector or a scalar, the idea being that internally I'd work with a vector. I had thought about passing a vector from one layer of code to the other, and I still might using a left argument to control code. With I pass the top level a vector the code runs enormously faster, so my guess is that most of the cpu is being eaten by converting very long numbers from vectors to digits. The recursive routine always passes down a vector when it recurs which might be why it is almost as fast as the loop.
That does not change my question.
I have an answer for this which I can't post for three hours. I will post it then, please don't do a ton of research to answer it.
assignments like
arr=. 'z' 15} arr
are executed in place. (See JWiki article for other supported in-place operations)
Interpreter determines that only small portion of arr is updated and does not create entire new list to reassign.
What happens in your case is not that array is being reassigned, but that it grows many times in small increments, causing memory allocation and reallocation.
If you preallocate (by assigning it some large chunk of data), then you can modify it with } without too much penalty.
After I asked this question, to be honest, I lost track of this web site.
Yes, the answer is that the language has no form that means "update in place, but if you use two forms
x =: x , most anything
or
x =: most anything } x
then the interpreter recognizes those as special and does update in place unless it can't. There are a number of other specials recognized by the interpreter, like:
199(1000&|#^)199
That combined operation is modular exponentiation. It never calculates the whole exponentiation, as
199(1000&|^)199
would - that just ends as _ without the #.
So it is worth reading the article on specials. I will mark someone else's answer up.
The link that sverre provided above ( http://www.jsoftware.com/jwiki/Essays/In-Place%20Operations ) shows the various operations that support modifying an existing array rather than creating a new one. They include:
myarray=: myarray,'blah'
If you are interested in a tacit version of the lookandsay sequence see this submission to RosettaCode:
las=: ,#((# , {.);.1~ 1 , 2 ~:/\ ])&.(10x&#.inv)#]^:(1+i.#[)
5 las 1
11 21 1211 111221 312211